Binet's Formula with Cosines

Elsewhere we found cosines and sines of several angles related to the regular pentagon.

\(\mbox{cos}(36^{\circ}) = \frac{(1 + \sqrt{5})}{4}.\)

For the following I shall need one more:

\(\mbox{cos}(108^{\circ}) = -\mbox{cos}(72^{\circ})=-\mbox{sin}(18^{\circ})=\frac{(1 - \sqrt{5})}{4}.\)

Observe that \(36^{\circ} = \frac{\pi}{5}\) and \(108^{\circ} = \frac{3\pi}{5}\). On the other hand, \(\frac{(1 + \sqrt{5})}{4}=\frac{\phi}{2}\), half of the Golden Ratio, while \(\frac{(1 - \sqrt{5})}{4}=\frac{1}{2}[-\frac{1}{\phi}]\).

What I plan to do is to combine those values to rewrite the Binet's formula for the Fibonacci numbers:

\(F_{n}=\frac{1}{\sqrt{5}}({\phi}^{n}-{\tau}^{n}),\)

where \(\tau\) is exactly \(-\frac{1}{\phi}\).

The combination of the above obtained, according to [Grimaldi, p. 67], first by W. Hope-Jones in 1921, leads to

\(F_{n}=\frac{1}{\sqrt{5}}(2^{n})\left[\mbox{cos}^{n}(\frac{\pi}{5})-\mbox{cos}^{n}(\frac{3\pi}{5})\right].\)

References

R. Grimaldi, Fibonacci and Catalan Numbers: an Introduction, Wiley, 2012

Trigonometry

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